Chapter 1: Intersection Theory on Surfaces

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Math

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https://git.sr.ht/~skylermarks/hartshorne

Exercises from Section V, Chapter 1 of Hartshorne’s Algebraic Geometry; Intersection Theory on Surfaces. Surfaces are nonsingular and projective. Curves are effective divisors on surfaces.

Exercise V.1.2:

Let $H$ be a very ample divisor on the surface $X$ , corresponding to a projective embedding $X\subset \mb P^n$ . If we write the Hilbert polynomial of $X$ as

P(z) = \frac12az^2+bz+c

Show that

a= H^2, b = \frac12 H^2+1-\pi, c=1+p_a

Where $\pi$ is the genus of a nonsingular curve linearly equivalent to $H$ (which exists by Bertini). Moreover, show that if $C$ is any curve in $X$ , then the degree of $C$ under the same embedding in $\mb P^n$ is exactly $C\cdot H$ .

Solution. Note that the Riemann Roch formula gives for any multiple of $H$

\dim_kH^0(X, \mc O_X(nH)) - \dim_kH^1(X, \mc O_X(nH)) + \dim_kH^0(X, \mc O_X(K+nH)) = \frac 12 nH\cdot(nH-K)+1+p_a

But by Serre duality, $H^0(X, \mc O_X(K+nH))\cong H^2(X, \mc O_X(nH))$ , so we have

\chi(nH) = \frac 12 nH\cdot(nH-K)+1+p_a = P(n)

Because the Hilbert polynomial is the polynomial which assigns to $n$ the Euler characteristic of the $n$ th twist of the sheaf defining the embedding. Then we have

P(n) = \frac 12 n^2H^2 - \frac12 H\cdot K+1+p_a

But, replacing $H$ with a nonsingular curve linearly equivalent to $H$ (since we work only up to linear equivalence), the adjunction formula gives

H\cdot(H-K) = 2g-2

H^2-2\pi+2 = H\cdot K

P(n) = \frac 12 n^2H^2 - \frac12 (H^2-2g+2)n+1+p_a

and we are done with the first part.

For the second part, recall that a very ample divisor is a divisor associated to the pullback of $\mc O(1)$ on $\mb P^n$ for some embedding $i$ , and so if $C$ is a curve which embedds via $j$ into $X$ , then $i\circ j$ is an embedding into $\mb P^n$ and $j^ * \mc L(H) = j^*i^*\mc O(1)$ gives rise to a very ample divisor $j^* H$ on $C$ . Then we have by Riemann-Roch for curves and Serre duality

P(n) = \chi(nj^*i*\mc O(1)) = \deg(i^* nH) -g +1

And so it suffices to show that $\deg_C i^* H = C\cdot H$ . Choosing a curve $H'$ which meets $C$ transversely and is linearly equivalent to $H$ without loss of generality (by a corollary of Bertini), we may write $i^* \mc O(-H)$ as the locally free sheaf defined by $i^\sharp(f_i)$ on $U_i\times_XC$ , where $\mc O(-H)$ is generated on $U_i$ by $f_i$ for an open cover $\{U_i\}$ . Then using the intersection multiplicity definition of the intersection pairing for effective divisors yields the intended consequence, and we are done.